Problem: $f(x) = \dfrac{ 6 }{ \sqrt{ 3 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $3 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 3$ , which means $-3 \leq x \leq 3$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 3 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 3$ This means that $x \neq 3$ and $x \neq -3$ So we have four restrictions: $x \geq -3$ $x \leq 3$ $x \neq -3$ , and $x \neq 3$ Combining these four, we know that $x > -3$ and $x < 3$ ; alternatively, that $-3 < x < 3$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -3< x <3\, \}$.